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Let $G_0$ be a finitely presented infinite group which is generated by a finite number of elements of order $3$.

Let $x$ be one of these elements and consider the normal closure $$G_n:= \langle x^{G_{n-1}}\rangle$$ of $x$ in $G_{n-1},$ $n\geq 1$.

Assume that $|G_0:G_1|=3, |G_1:G_2|=3$.

What are the possibilities for $|G_{n-1}:G_n|,$ $n\geq 1$? Can it be $>3$? Can it be $\infty$? Under what conditions can we conclude that it is always $3$, or finite?

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1 Answer 1

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$|G_2:G_3|$ can be an arbitrarily large finite number (although I don't know exactly what values it could take). If there is a finitely presented infinite simple group generated by elements of order $3$, then it can also be infinite.

I found an example of a group $P$ of order $3^4$ satisfying your hypotheses, but that has $|P_2:P_3|=3$,

If we take a wreath product $W$ of a simple group $S$ that can be generated by elements of order $3$ with $P$ in a degree $27$ permuation representation, then $|W_0:W_1|=|W_1:W_2|=3$, but $|W_2:W_3| = 3|S|^3$. This is verified for $S=A_5$ in the Magma calculation that follows.

> F:=Group<x,y|x^3,y^3>;
> P:=pQuotient(F,3,3);
> P:=quo<P|P.5>;
> #P;
81
> Q:=CosetImage(P,sub<P|P.1>);
> W:=WreathProduct(Alt(5),Q);
> [Order(W.i): i in [1..Ngens(W)]];
[ 3, 3, 3, 3, 3, 3 ]
> N:=ncl<W|W.1>;
> Index(W,N);
3
> N2:=ncl<N|W.1>;
> Index(N,N2);
3
> N3:=ncl<N2|W.1>;
> Index(N2,N3);
648000
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  • $\begingroup$ Thank you for your answer.I forgot to say that my group is infinite. I think that there is no such simple subgroup. Do you think that the index could be infinite anyway? $\endgroup$
    – CRito
    Commented Oct 26, 2016 at 14:54
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    $\begingroup$ If $G$ is a simple group with an element of order 3, then clearly it's generated by its elements of order 3. Thompson's groups $T$ and $V$ (on the circle and Cantor set) are such groups and are finitely presented. $\endgroup$
    – YCor
    Commented Oct 26, 2016 at 15:07
  • $\begingroup$ Actually there is no need for the group $S$ in my example to be generated by elements of order $3$. You can generate the group with a finite number of conjugates of the generators of $P$ in $W$. $\endgroup$
    – Derek Holt
    Commented Oct 26, 2016 at 15:54

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